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Question Number 118043 by prakash jain last updated on 14/Oct/20

$$\mathrm{How}\mathrm{many}\mathrm{numbers}\mathrm{in}\mathrm{the}\mathrm{range}$$$$[1,{10}^n-1]\mathrm{are}\mathrm{divisible}\mathrm{by}9?$$$$\mathrm{digit}\mathrm{repetitions}\mathrm{are}\mathrm{not}\mathrm{allowed}.$$

Commented by mr W last updated on 15/Oct/20

$$withoutrepetitionmeansmax.$$$$10digitsareallowed,i.e.n⩽10.$$$$$$$$10digitnumbers:$$$$0+1+2+..+9=45✓$$$$\Rightarrow \frac{9}{10}\times 10!=9\times 9!$$$$$$$$9digitnumbers:$$$$1+2+..+9=45✓$$$$\Rightarrow 9!$$$$0+1+2+..+8=36✓$$$$\Rightarrow \frac{8}{9}\times 9!=8\times 8!$$$$\Rightarrow 9!+8\times 8!=17\times 8!$$$$$$$$8digitnumbers:$$$$without0+9\Rightarrow 8!$$$$without1+8,2+7,3+6,4+5\Rightarrow 4\times \frac{7}{8}\times 8!=4\times 7\times 7!$$$$\Rightarrow 8!+4\times 7\times 7!=36\times 7!$$$$$$$$7digitnumbers:$$$$without0+1+8,0+2+7,0+3+6,0+4+5\Rightarrow 4\times 7!$$$$without1+8+9,2+7+9,3+6+9,4+5+9\Rightarrow 4\times \frac{6}{7}\times 7!=4\times 6\times 6!$$$$without1+2+6,1+3+5\Rightarrow 2\times \frac{6}{7}\times 7!=2\times 6\times 6!$$$$without2+3+4\Rightarrow \frac{6}{7}\times 7!=6\times 6!$$$$\Rightarrow (4\times 7+4\times 6+2\times 6+6)6!=10\times 7!$$$$$$$$6digitnumbers:$$$$without0+1+2+6,0+1+3+5,0+2+3+4\Rightarrow 3\times 6!$$$$without1+2+6+8,1+3+5+9,2+3+4+9\Rightarrow 3\times \frac{5}{6}\times 6!=3\times 5\times 5!$$$$\Rightarrow 3(6+5)5!=33\times 5!$$$$$$$$5digitnumbers:$$$$9+8+7+2+1$$$$9+8+6+3+1$$$$9+8+5+4+1$$$$9+8+5+3+2$$$$9+7+6+4+1$$$$9+7+6+3+2$$$$9+7+5+4+2$$$$9+6+5+4+3$$$$$$$$......$$

Commented by prakash jain last updated on 15/Oct/20

$$\mathrm{Thanks}\mathrm{sir}.$$$$\mathrm{It}\mathrm{looks}\mathrm{listing}\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{possible}\mathrm{way}.$$$$\mathrm{I}\mathrm{thought}\mathrm{I}\mathrm{had}\mathrm{seen}\mathrm{a}\mathrm{closed}\mathrm{forum}$$$$\mathrm{expression}\mathrm{somewhere}\mathrm{but}\mathrm{was}\mathrm{not}$$$$\mathrm{able}\mathrm{to}\mathrm{derive}.$$

Answered by floor(10²Eta[1]) last updated on 24/Dec/20

$$\mathrm{from}1\mathrm{to}{10}^{\mathrm{n}}-1\mathrm{we}\mathrm{have}\frac{{10}^{\mathrm{n}}-1}{9}\mathrm{multiples}\mathrm{of}9$$$$\left(\mathrm{considering}\mathrm{the}\mathrm{numbers}\mathrm{with}\mathrm{the}\mathrm{same}\mathrm{digits}\right)$$$$\mathrm{so}{\mathrm{let}}^{\prime }\mathrm{s}\mathrm{calculate}\mathrm{how}\mathrm{many}\mathrm{numbers}\mathrm{have}$$$$\mathrm{the}\mathrm{same}\mathrm{digits}\mathrm{and}\mathrm{are}\mathrm{divisible}\mathrm{by}9$$$$[\mathrm{aa}...\mathrm{a}]\left(\mathrm{n}\mathrm{digits}\right)$$$$\mathrm{a}.\mathrm{n}\equiv 0\left(\mathrm{mod}9\right),\mathrm{a}\in [1,9]\wedge \mathrm{n}⩾2$$$$\mathrm{a}\in \{1,2,4,5,7,8\}\Rightarrow \mathrm{n}=9\mathrm{k},\mathrm{k}⩾1$$$$\mathrm{a}\in \{3,6\}\Rightarrow \mathrm{n}=3\mathrm{k},\mathrm{k}⩾1$$$$\mathrm{a}=9\Rightarrow \mathrm{n}=\mathrm{k}⩾2$$$$2\mathrm{digits}:\left(99\right)$$$$3\mathrm{digits}:(333,666,999)$$$$4\mathrm{digits}:\left(9999\right)$$$$5\mathrm{digits}:\left(99999\right)$$$$6\mathrm{digits}:(333333,666666,999999)$$$$[...]$$$$9\mathrm{digits}:(11...1,22...2,33...3,...,99..9)$$$$2\mathrm{digits}\mathrm{to}9\mathrm{digits}:5+3.2+9=20\mathrm{cases}$$$$10\mathrm{digits}\mathrm{to}19:7+3.2+9=22\mathrm{cases}$$$$\mathrm{from}1\mathrm{to}\mathrm{x}\mathrm{we}\mathrm{have}$$$$\mathrm{x}-\lfloor \frac{\mathrm{x}}{3}\rfloor \mathrm{numbers}\mathrm{that}\mathrm{are}\mathrm{coprime}\mathrm{with}3$$$$(\mathrm{because}\lfloor \frac{\mathrm{x}}{3}\rfloor \mathrm{is}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{multiples}\mathrm{of}3,\mathrm{from}1\mathrm{to}\mathrm{x})$$$$\mathrm{from}2\mathrm{to}\mathrm{x}\mathrm{we}\mathrm{have}$$$$(\mathrm{x}-\lfloor \frac{\mathrm{x}}{3}\rfloor -1)+3(\lfloor \frac{\mathrm{x}}{3}\rfloor -\lfloor \frac{\mathrm{x}}{9}\rfloor )+9\lfloor \frac{\mathrm{x}}{9}\rfloor$$$$\mathrm{x}-1+2\lfloor \frac{\mathrm{x}}{3}\rfloor +6\lfloor \frac{\mathrm{x}}{9}\rfloor \mathrm{numbers}\mathrm{that}\mathrm{are}\mathrm{divisible}$$$$\mathrm{by}9\mathrm{and}\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{digits}$$$$\mathrm{from}2\mathrm{digits}\mathrm{to}\mathrm{n}\mathrm{digits}\mathrm{we}\mathrm{have}$$$$\mathrm{n}-1+2\lfloor \frac{\mathrm{n}}{3}\rfloor +6\lfloor \frac{\mathrm{n}}{9}\rfloor \mathrm{numbers}\mathrm{that}\mathrm{have}$$$$\mathrm{the}\mathrm{same}\mathrm{digits}\mathrm{and}\mathrm{are}\mathrm{divisible}\mathrm{by}9$$$$$$$$\mathrm{Ans}:$$$$\frac{{10}^{\mathrm{n}}-1}{9}-(\mathrm{n}-1+2\lfloor \frac{\mathrm{n}}{3}\rfloor +6\lfloor \frac{\mathrm{n}}{9}\rfloor )$$$$$$

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